Captain Brandon has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Daniel and his merciless band of thieves. The Captain has probability $\dfrac{3}{8}$ of hitting the pirate ship, if his ship hasn't already been hit. If it has been hit, he will always miss. The pirate has probability $\dfrac{1}{3}$ of hitting the Captain's ship, if his ship hasn't already been hit. If it has been hit, he will always miss as well. If the pirate shoots first, what is the probability that both the pirate and the Captain hit each other's ships?
Answer: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is the pirate hitting the Captain's ship and event B is the Captain hitting the pirate ship. The pirate fires first, so his ship can't be sunk before he fires his cannons. So, the probability of the pirate hitting the Captain's ship is $\dfrac{1}{3}$ If the pirate hit the Captain's ship, the Captain has no chance of firing back. So, the probability of the Captain hitting the pirate ship given the pirate hitting the Captain's ship is $0$ The probability that both the pirate and the Captain hit each other's ships is then the probability of the pirate hitting the Captain's ship times the probability of the Captain hitting the pirate ship given the pirate hitting the Captain's ship. This is $\dfrac{1}{3} \cdot 0 = 0$